No. For every infinite linear order $L$ and every free ultrafilter $\mathcal{U}$ on $\omega$, the ultrapower $L^\omega{/}\mathcal{U}$ is $\aleph_0$-saturated. And infinite $\aleph_0$-saturated linear orders cannot be complete!

Proof: Let $L$ be an infinite $\aleph_0$-saturated linear order. Since $L$ is infinite, it contains an infinite increasing sequence or an infinite decreasing sequence. Without loss of generality, let's say we have an increasing sequence $(a_n)_{n\in\omega}$.

By $\aleph_0$-saturation, the partial type $\{x>a_n\mid n\in\omega\}$ is realized in $L$, so the set $\{a_n\mid n\in\omega\}$ is bounded above. Suppose $b$ is an upper bound. Then the partial type $\{x>a_n\mid n\in\omega\}\cup \{x<b\}$ is realized in $L$, so $b$ is not a least upper bound. Thus $L$ is not complete.

No. Every ultraproduct by a free ultrafilter on $\omega$ is $\aleph_1$-saturated. And infinite $\aleph_1$-saturated linear orders cannot be complete!

Proof: Let $L$ be an infinite $\aleph_1$-saturated linear order. Since $L$ is infinite, it contains an infinite increasing sequence or an infinite decreasing sequence. Without loss of generality, let's say we have an increasing sequence $(a_n)_{n\in\omega}$.

By $\aleph_1$-saturation, the partial type $\{x>a_n\mid n\in\omega\}$ is realized in $L$, so the set $\{a_n\mid n\in\omega\}$ is bounded above. Suppose $b$ is an upper bound. Then the partial type $\{x>a_n\mid n\in\omega\}\cup \{x<b\}$ is realized in $L$, so $b$ is not a least upper bound. Thus $L$ is not complete.